Show that the series $\sum_{k=0} ^\infty (-1)^k \frac{x^{2k+1}}{2k+1}$ converges for $|x|<1$ and that it converges to $\arctan x$ Show that the series $\sum_{k=0} ^\infty (-1)^k \dfrac{x^{2k+1}}{2k+1}$ converges for $|x|<1$ and that it converges to $\arctan x$
I tried using the ratio test but I got that it equals 1, so it is inconclusive. How can I show that it converges, and that it converges to arctan x? I think I'm trying to show that the Legrange error term is 0 but I'm not sure how to proceed with it - the nth derivatives of arctan x keep changing so...
 A: Hint:  Start with
$$\frac1{1+x^2}=\sum_{n=0}^\infty (-1)^nx^{2n},\quad\text{for }\,|x|<1$$
and integrate both sides.
A: Let
$$u_k=(-1)^k \dfrac{x^{2k+1}}{2k+1}$$
then by the ratio test we have
$$\left|\frac{u_{k+1}}{u_k}\right|\xrightarrow{k\to\infty}|x|^2<1\iff |x|<1$$
hence the radius of convergence is $R=1$. We see easily using the Leibniz criteria that the domain of convergence is $[-1,1]$. Finally notice that
$$\arctan'(x)=\frac1{1+x^2}=\sum_{k=0}^\infty(-1)^k x^{2k}$$
so integrate term by term to get the series of $\arctan(x)$.
A: You know that this is a converging function because it is a sum of terms alternating in sign whose absolute values decrease and approach zero.  If you can prove the terms alternate in sign and their absolute values decrease and approach zero, then it follows that the series converges.  
See this question for why it goes to arctan: Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$?
A: Assuming $|x|<1$,
$$ \sum_{k=0}^{N}(-1)^k\frac{x^{2k+1}}{2k+1} = \sum_{k=0}^{N}(-1)^k\int_{0}^{x}y^{2k}\,dy = \int_{0}^{x}\frac{1-(-y^2)^{N+1}}{1+y^2}\,dy$$
where:
$$\int_{0}^{x}\frac{dy}{1+y^2} = \arctan x$$
and:
$$\left|\int_{0}^{x}\frac{(-y^2)^{N+1}}{1+y^2}\,dy\right|\leq \int_{0}^{|x|}y^{2N+2}\,dy=\frac{|x|^{2N+3}}{2N+3}\to 0.$$
A: Modifying Jack D'Aurizio's answer,
since
$\sum_{k=0}^n x^k = \dfrac{1-x^{n+1}}{1-x}$,
putting $-x^2$ for $x$ we get
$\sum_{k=0}^n (-1)^kx^{2k} 
= \dfrac{1-(-1)^{n+1}x^{2n+2}}{1+x^2}
= \dfrac1{1+x^2}+\dfrac{(-1)^{n}x^{2n+2}}{1+x^2}
$
or
$\dfrac1{1+x^2}
=\sum_{k=0}^n (-1)^kx^{2k} -\dfrac{(-1)^{n}x^{2n+2}}{1+x^2}
$.
Integrating from $0$ to $y$,
$\begin{array}\\
\arctan(y)
&=\int_0^y \dfrac{dx}{1+x^2}\\
&=\int_0^y \left(\sum_{k=0}^n (-1)^{k-1}x^{2k}-(-1)^{n}\dfrac{x^{2n+2}}{1+x^2}\right)dx\\
&=\sum_{k=0}^n (-1)^{k-1}\int_0^y x^{2k}dx-(-1)^{n}\int_0^y \dfrac{x^{2n+2}dx}{1+x^2}\\
&=\sum_{k=0}^n \dfrac{(-1)^{k-1}x^{2k+1}}{2k+1}-(-1)^{n}\int_0^y \dfrac{x^{2n+2}dx}{1+x^2}\\
\end{array}
$
so
$\arctan(y)-\sum_{k=0}^n \dfrac{(-1)^{k-1}x^{2k+1}}{2k+1}
=(-1)^{n+1}\int_0^y \dfrac{x^{2n+2}dx}{1+x^2}
$
so that
$\begin{array}\\
|\arctan(y)-\sum_{k=0}^n \dfrac{(-1)^{k-1}x^{2k+1}}{2k+1}|
&\le\int_0^y \dfrac{x^{2n+2}dx}{1+x^2}\\
&\le\int_0^y x^{2n+2}dx\\
&=\dfrac{x^{2n+3}}{2n+3}\\
&\to 0
\qquad\text{as } n \to \infty\\
\end{array}
$
